JEE MAIN - Physics (2020 - 6th September Morning Slot - No. 14)
If the potential energy between two molecules is given by
U = $$ - {A \over {{r^6}}} + {B \over {{r^{12}}}}$$,
then at equilibrium, separation between molecules, and the potential energy are :
U = $$ - {A \over {{r^6}}} + {B \over {{r^{12}}}}$$,
then at equilibrium, separation between molecules, and the potential energy are :
$${\left( {{{2B} \over A}} \right)^{1/6}}$$, $$ - {{{A^2}} \over {4B}}$$
$${\left( {{{2B} \over A}} \right)^{1/6}}, - {{{A^2}} \over {2B}}$$
$${\left( {{B \over A}} \right)^{1/6}},0$$
$${\left( {{B \over {2A}}} \right)^{1/6}}, - {{{A^2}} \over {2B}}$$
Explanation
U = $$ - {A \over {{r^6}}} + {B \over {{r^{12}}}}$$
F = - $${{dU} \over {dr}}$$
= – (A(–6r–7 )) + B(–12r–13)
for equilibrium, F = 0
$$ \therefore $$ 0 = $${{6A} \over {{r^7}}} - {{12B} \over {{r^{13}}}}$$
$$ \Rightarrow $$ $${{6A} \over {12B}} = {1 \over {{r^6}}}$$
$$ \Rightarrow $$ r = $${\left( {{{2B} \over A}} \right)^{{1 \over 6}}}$$
$$ \therefore $$ U = $$ - {A \over {{{2B} \over A}}} + {B \over {{{\left( {{{2B} \over A}} \right)}^2}}}$$
= $$ - {{{A^2}} \over {2B}} + {{{A^2}} \over {4B}}$$
= $$ - {{{A^2}} \over {4B}}$$
F = - $${{dU} \over {dr}}$$
= – (A(–6r–7 )) + B(–12r–13)
for equilibrium, F = 0
$$ \therefore $$ 0 = $${{6A} \over {{r^7}}} - {{12B} \over {{r^{13}}}}$$
$$ \Rightarrow $$ $${{6A} \over {12B}} = {1 \over {{r^6}}}$$
$$ \Rightarrow $$ r = $${\left( {{{2B} \over A}} \right)^{{1 \over 6}}}$$
$$ \therefore $$ U = $$ - {A \over {{{2B} \over A}}} + {B \over {{{\left( {{{2B} \over A}} \right)}^2}}}$$
= $$ - {{{A^2}} \over {2B}} + {{{A^2}} \over {4B}}$$
= $$ - {{{A^2}} \over {4B}}$$
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